\hfill \thepage} %} \geometry{left=0.6in,right=0.7in,top=0.7in,bottom=0.9in} \pagestyle{fancy} \lhead{\Large \color{blue}Lecture Notes} \lfoot{} \cfoot{} \chead{\LARGE Area} \rhead{\Large page \thepage} \textwidth 7.3in \textheight 3in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \input{tcilatex} \begin{document} \textbf{Definition: }The \textbf{area} of a geometric object is a measurement of its surface.\bigskip While we could think about perimeter as a fencing problem, area can be thought of as follows. Suppose a geometric object is a room. How much rug do we need to buy to cover the entire room? \ Understanding and remembering the area formulas are probably easier if we know how they were derived.\bigskip \textbf{Definition: \ }The area of a $1$ feet by $1$ feet square is defined to be $1$ ft$^{2}$ $\ $(square-feet)$.$ (Similar definitions can be formulated with mi$^{2},$ cm$^{2},$ in$^{2},$ etc. ) \ The area of an object, measured in ft$^{2},$ is the number of $1$ ft by $1$ ft square needed to cover the object, cutting and pasting allowed.\FRAME{dtbpF}{% 0.9392in}{0.9124in}{0pt}{}{}{insert1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.9392in;height 0.9124in;depth 0pt;original-width 0.9063in;original-height 0.8804in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert1.bmp';file-properties "XNPEU";}} Area is not a length like perimeter. Area is always measured in ft$^{2},$ mi$% ^{2},$ cm$^{2},$ in$^{2},$ etc., and is usually denoted by $A$. \ \textbf{Theorem: }The area of a rectangle with sides $a$ and $b$ is $A=ab$. Proof: \ Consider rectangle with sides $3$ m and $5$ m. \ The area of this rectangle will be as many m$^{2}$ as many $1$ m by $1$ m square is needed to cover it. Once we place this grid on the rectangle, it is easy to see, just how many squares we need. \FRAME{dtbpF}{1.5022in}{0.9928in}{0pt}{}{}{insert2.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.5022in;height 0.9928in;depth 0pt;original-width 1.4667in;original-height 0.9599in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert2.bmp';file-properties "XNPEU";}} We used exactly $15$ squares to cover the rectangle, and so the area is $15$ m$^{2}$. Mathematicians also proved that the formula is true even if the sides of the rectangle are not integers. It is interesting to see that we basically counted how many meter$^{2}$ we have. A computation for the area, including the units is slightly different. Instead of counting meter$^{2},$ we literally multiply meter by meter. \begin{equation*} A=ab=3\text{ m }\left( 5\text{ m}\right) =15\text{ m}^{2} \end{equation*} Area computation will always yield the right unit. \begin{enumerate} \item Find the area of a rectangle with sides $13$ in and $7$ in. Solution: \ We apply the formula.$A=ab$.% \begin{equation*} A=ab=13\text{ in }\left( 7\text{ in}\right) =91\text{ in}^{2} \end{equation*}% The following few area formulas will demonstrate how mathematicians work: we will use already proven results to come up with new formulas.\pagebreak \textbf{Theorem: }The area of a right triangle with sides $a$, $b$, and $c$ (where $c$ is the longest side) is $A=\dfrac{ab}{2}$.\FRAME{dtbpF}{2.0384in}{% 1.0741in}{0in}{}{}{insert3.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.0384in;height 1.0741in;depth 0in;original-width 2.0003in;original-height 1.0395in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert3.bmp';file-properties "XNPEU";}}Proof: \ It is very easy to see that every right triangle is basically half of a rectangle. We can make a rectangle if we use two identical right triangles as shown on the picture below.\FRAME{dtbpF}{1.8706in}{1.2816in}{0pt}{}{}{insert3.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.8706in;height 1.2816in;depth 0pt;original-width 1.8334in;original-height 1.2462in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert3.bmp';file-properties "NPEU";}}Since the rectangle's sides are $a$ and $b$, its area is $ab.$ The area of our triangle must be half of it. Thus% \begin{equation*} A=\dfrac{ab}{2} \end{equation*}% Notice that we never used the length of the longest side, $c$. \item Find the area of the right triangle with sides $5$ mi, $12$ mi, and $% 13 $ mi long. Solution: \ It is important to know that the largest side, $13$ mi long, is not needed for this computation. With labeling $a=5$ mi and $b=12$ mi, the area is% \begin{equation*} A=\dfrac{ab}{2}=\dfrac{5\text{ mi }\left( 12\text{ mi}\right) }{2}=30\text{ mi}^{2} \end{equation*}% \textbf{Theorem:} \ The area of a general triangle with sides $a,$ $b,$ $c$ and height $h$ as shown on the picture below is $A=\dfrac{ah}{2}$.\FRAME{% dtbpF}{1.5091in}{1.0871in}{0pt}{}{}{insert4.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.5091in;height 1.0871in;depth 0pt;original-width 1.4728in;original-height 1.0533in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert4.bmp';file-properties "NPEU";}}Proof: \ As before, we will use a previously obtained result. Since the general triangle no longer has a right angle, we create it by drawing in the altitude or height belonging to the side $a.$ Now we split our triangle into two right triangles, and each of them is half of a rectangle.\FRAME{% dtbpF}{1.8706in}{1.3949in}{0pt}{}{}{insert5.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.8706in;height 1.3949in;depth 0pt;original-width 1.8334in;original-height 1.3604in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert5.bmp';file-properties "NPEU";}}Our triangle makes up for half of a rectangle, with sides $a$ and $% h $. Thus $A=\dfrac{ah}{2}$. \item Find the area of the triangle shown on the picture below.\FRAME{dtbpF}{% 1.8706in}{1.3949in}{0pt}{}{}{insert5b.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.8706in;height 1.3949in;depth 0pt;original-width 1.8334in;original-height 1.3604in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert5b.bmp';file-properties "NPEU";}}Solution: It is important to notice that we will not need all the information given. We apply the area-formula.% \begin{equation*} A=\dfrac{ah}{2}=\dfrac{21\text{ in }\left( 12\text{ in}\right) }{2}=126\text{ in}^{2} \end{equation*}% \textbf{Definition: }\ A parallelogram is a four sided polygon with two pairs of parallel sides.\FRAME{dtbpF}{1.6034in}{0.966in}{0pt}{}{}{insert6.bmp% }{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.6034in;height 0.966in;depth 0pt;original-width 1.567in;original-height 0.9331in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert6.bmp';file-properties "NPEU";}}It is a proven fact that the opposite sides of a parallelogram are of equal length. \textbf{Theorem:\ }The area of a parallelogram with sides $a,$ $b$ and height $h$ belonging to $a$ is $A=ah$.\FRAME{dtbpF}{1.6034in}{0.966in}{0pt}{% }{}{insert7.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.6034in;height 0.966in;depth 0pt;original-width 1.567in;original-height 0.9331in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert7.bmp';file-properties "NPEU";}}\pagebreak Proof: We will use (surprise, surprise!) a previously proven result. If we cut off a triangle and paste it back as show on the picture below, we obtain a rectangle.\FRAME{dtbpF}{3.8951in}{0.966in}{0pt}{}{}{insert8.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.8951in;height 0.966in;depth 0pt;original-width 3.8467in;original-height 0.9331in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert8.bmp';file-properties "NPEU";}}Thus the area of the parallelogram equals to the area of a rectangle with sides $a$ and $h$. \item Find the area of the parallelogram shown on the picture below.\FRAME{% dtbpF}{1.7772in}{0.9591in}{0pt}{}{}{insert9.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.7772in;height 0.9591in;depth 0pt;original-width 1.74in;original-height 0.9271in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert9.bmp';file-properties "NPEU";}}Solution: \ We apply the formula for the area of a trapezoid..% \begin{equation*} A=ah=5\text{ cm }\left( 3\text{ cm}\right) =15\text{ cm}^{2} \end{equation*}% \textbf{Definition:} \ A \textbf{trapezoid} is a four sided polygon with one pair of parallel sides.\FRAME{dtbpF}{1.6034in}{0.966in}{0pt}{}{}{insert10.bmp% }{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.6034in;height 0.966in;depth 0pt;original-width 1.567in;original-height 0.9331in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert10.bmp';file-properties "NPEU";}}\textbf{Theorem: }The area of a trapezoid, with sides and height labeled as on the picture below, is $A=\dfrac{a+b}{2}h$.\FRAME{dtbpF}{% 1.6034in}{0.966in}{0pt}{}{}{insert11.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.6034in;height 0.966in;depth 0pt;original-width 1.567in;original-height 0.9331in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert11.bmp';file-properties "NPEU";}}\pagebreak Proof: \ If we use two identical trapezoids, we can make a parallelogram as shown on the picture below. \FRAME{dtbpF}{2.6411in}{0.966in}{0pt}{}{}{% insert12.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.6411in;height 0.966in;depth 0pt;original-width 2.5996in;original-height 0.9331in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert12.bmp';file-properties "NPEU";}}We already know that the area of this parallelogram is $A=\left( a+b\right) h$. Since our trapezoid is exactly half of the parallelogram, its area is $A=\dfrac{\left( a+b\right) h}{2}$. \item Find the area of the trapezoid shown on the picture below.\FRAME{dtbpF% }{1.4088in}{0.966in}{0pt}{}{}{insert13.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.4088in;height 0.966in;depth 0pt;original-width 1.3733in;original-height 0.9331in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert13.bmp';file-properties "NPEU";}}We apply the formula% \begin{equation*} A=\dfrac{a+b}{2}h=\dfrac{3\text{ cm }+5\text{ cm}}{2}\left( 2\text{ cm}% \right) =\dfrac{8\text{ cm}}{2}\left( 2\text{ cm}\right) =8\text{ cm}^{2} \end{equation*}% \textbf{Definition: }The area of a circle with radius $r$ is $A=\pi r^{2}$. At this level of mathematics, we do not have the tools necessary to prove this. \ The proof actually uses (again) the formula for the area of a rectangle. \ Instead of a circle, we use just half of it, and then we multiply the result by $2$. \ The basic idea is to first find an estimation for the area, using rectangles.\FRAME{dtbpF}{1.6743in}{1.0836in}{0pt}{}{}{% insert14.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.6743in;height 1.0836in;depth 0pt;original-width 1.2194in;original-height 0.7801in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert14.bmp';file-properties "NPEU";}}We CAN compute the grey area since it is composed of rectangles. \ But this is a crude underestimation of the actual area. The trick is that the more rectangles we use, the more accurate the approximation becomes. \ Using more and more rectangles to estimate the area of the semi-circle, the approximation becomes better and better.\FRAME{dtbpF}{2.0496in}{1.3266in}{0pt% }{}{}{insert15.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.0496in;height 1.3266in;depth 0pt;original-width 1.2194in;original-height 0.7801in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert15.bmp';file-properties "NPEU";}}Calculus offers tools to find a unique number these approximations approach. This number is the area. \item Find the area of a circle of radius $7$ ft. Solution: \ We apply the formula.% \begin{equation*} A=\pi r^{2}=\pi \left( 7\text{ ft}\right) ^{2}=153.\,\allowbreak 94\text{ ft}% ^{2} \end{equation*} \item Find the area of the figure shown on the picture below.\FRAME{dtbpF}{% 2.5097in}{2.06in}{0pt}{}{}{insert16.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.5097in;height 2.06in;depth 0pt;original-width 2.469in;original-height 2.0211in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra2_lnotes/area/insert16.bmp';file-properties "NPEU";}}Solution: \ The area is the sum of that of a rectangle and a semi-circle. We apply the appropriate formulas and then add the results. The bottom side of the rectangle is $6$ cm since we can lay exactly two radii on it.% \begin{eqnarray*} A_{\text{rectangle}} &=&ab=4\text{ cm }\left( 6\text{ cm}\right) =24\text{ cm% }^{2} \\ A_{\text{semicircle}} &=&\dfrac{\pi r^{2}}{2}=\dfrac{\pi \left( 3\text{ cm}% \right) ^{2}}{2}=14.\,\allowbreak 137\text{ cm}^{2} \\ A &=&A_{\text{rectangle}}+A_{\text{semicircle}}=24\text{ cm}% ^{2}+14.\,\allowbreak 137\text{ cm}^{2}=38.\,\allowbreak 137\text{ cm}^{2} \end{eqnarray*} \end{enumerate} \pagebreak {\Large Practice Problems}\bigskip \begin{enumerate} \item The sides of a rectangle with sides $4~\unit{in}$ and $11~\unit{in}$. \ Find the area of the rectangle. \ Include units in your answer. \item Find the area of the triangle.shown on the picture below.~ \FRAME{itbpF}{2.4682in}{1.1675in}{-0.07in}{}{}{insert7.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.4682in;height 1.1675in;depth -0.07in;original-width 2.4267in;original-height 1.1338in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra1_problemsets/ps07/insert7.bmp';file-properties "NPEU";}} \item Find the area of the figure shown on the picture below. \ Include units in your answer. \FRAME{itbpF}{2.5676in}{1.2142in}{0.01in}{}{}{insert.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.5676in;height 1.2142in;depth 0.01in;original-width 2.527in;original-height 1.1796in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra1_problemsets/ps11/insert.bmp';file-properties "NPEU";}} \item Find the area of the figure shown on the picture below. \ Include units in your answer. \FRAME{itbpF}{1.8645in}{1.4823in}{0.01in}{}{}{insert.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.8645in;height 1.4823in;depth 0.01in;original-width 1.8265in;original-height 1.4468in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra1_problemsets/ps12/insert.bmp';file-properties "NPEU";}} \item Find the area of a circle with radius $7~\unit{in}$ long.\pagebreak \end{enumerate} {\Large Answers for Practice Problems}\bigskip \begin{enumerate} \item The sides of a rectangle with sides $4~\unit{in}$ and $11~\unit{in}$. \ Find the area of the rectangle. \ Include units in your answer.~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion A=44~\unit{in}^{2}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \item Find the area of the triangle.shown on the picture below.~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion A=30$~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion \unit{cm}^{2}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \FRAME{itbpF}{2.4682in}{1.1675in}{-0.07in}{}{}{insert7.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.4682in;height 1.1675in;depth -0.07in;original-width 2.4267in;original-height 1.1338in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra1_problemsets/ps07/insert7.bmp';file-properties "NPEU";}} \item Find the area of the figure shown on the picture below. \ Include units in your answer.$~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion A=159~\unit{ft}^{2}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \FRAME{itbpF}{2.5676in}{1.2142in}{0.01in}{}{}{insert.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.5676in;height 1.2142in;depth 0.01in;original-width 2.527in;original-height 1.1796in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra1_problemsets/ps11/insert.bmp';file-properties "NPEU";}} \item Find the area of the figure shown on the picture below. \ Include units in your answer.~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion A=54~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion \unit{ft}^{2}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \FRAME{itbpF}{1.8645in}{1.4823in}{0.01in}{}{}{insert.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.8645in;height 1.4823in;depth 0.01in;original-width 1.8265in;original-height 1.4468in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../ALGEBRA/algebra1_problemsets/ps12/insert.bmp';file-properties "NPEU";}} \item Find the area of a circle with radius $7~\unit{in}$ long.~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion A=49\pi ~\unit{in}^{2}\cong 153.\,\allowbreak 938\,~\unit{in}^{2}$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \end{enumerate} \end{document}